2.1.24 Single Number II

Link: https://oj.leetcode.com/problems/single-number-ii/

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

我的思路：remember 應該用位操作。對於出現三次的元素，對應的位一定出現3次（還是3的倍數？）。每一位對3取餘，不爲0的位組合起來就是隻出現1次的數。

但我不知道通過什麼操作才能把一個數表示成二進制。//(A[j]>>i) & 1

Note： 沒法下筆。再做。

Time: O(n), Space: O(1) (O(32)

public class Solution {
    public int singleNumber(int[] A) {
        int[] count = new int[32];//store the total # of ones on each digit
        for(int i = 0; i < 32; i++){//for each digit
            for(int j = 0; j < A.length; j++){
                count[i] += (A[j]>>i) & 1;
            }
        }
        int result = 0;
        for(int i = 0; i < 32; i++){
            result += (count[i] %3) <<i;//can also use "|="
        }
        return result;
    }
}